Welcome to this tutorial on Exam C, problem 047. This problem falls under the learning objective "Construction & Selection of Parametric Models". Here's a view of the problem. Please complete the question and select one of the five answers. If incorrect, you can click the gray X to try again.
Here is a solution to the problem. This question asks us to use a chi-square test to determine whether the Poisson distribution is a good fit for the observed data on number of claims per day.
We will first estimate the Poisson parameter, lambda, using the method of maximum likelihood estimation. Then we will reorganize the data into four groups instead of five, which really doesn't actually take any work (the final group had no observations). Next, we can calculate the expected number of days in each category when once we have estimated the Poisson parameter. Then we'll have all the information we need to conduct the Chi-square test.
In order to estimate lambda, we can use a shortcut. Poisson is a distribution where maximum likelihood and method of moments both give the same answer, so we can calculate the sample mean to estimate the parameter. The steps to using MLE are not too complex though, so we'll do that too. We can first set up the likelihood equation as shown. Remember that constants can be disregarded since we will later be maximizing. Taking the natural log of the likelihood function, we get 600ln(lambda) minus 365 lambda. Next, taking the derivative and setting it equal to zero, we find a lambda estimate of 1.6438. Both methods match, so I'm feeling happy that that is the correct value.
We can estimate the expected probabilities now that we have a value of lambda. I am using N to denote the number of claims in a day. Use the Poisson distribution to find the probabilities for 0, 1 and 2. To find the probability of greater than or equal to 3, subtract the other probabilities from 1.
We need the expected number of days for each group, not just the probabilities. The sample contained a whole year of 365 days. Thus, we multiply 365 times each probability to find the expected number of days for each group. The results for each calculation are shown.
We're finally ready to perform the Chi-square test. To find the Chi-square statistic, we square the difference between the observed number and expected number, and divide by the expected. The summation of these yields the Chi-square statistic. The null hypothesis is that this data distribution is Poisson. The test will determine the level of certainty, or significance level, we have in saying we accept the null hypothesis, or whether we reject the null hypothesis entirely. It may be easier to perform this problem by making a table, as pictured.
It's time to decide whether to accept or reject the null hypothesis. To find degrees of freedom, we take the number of classes minus 1; however, we must also remove a degree of freedom for each parameter that we estimated using the sample data. We have four classes and estimated one parameter, so there are two degrees of freedom for this problem. We must check a Chi-square table, which is given as part of the inventory of distributions for the exam. The test statistic of 7.55 is greater than the critical value for 2.5% significance, but not 1% significance. We are 97.5% certain the data are not Poisson-distributed, but we cannot say we are 99% certain that we should reject the null hypothesis. This result corresponds with answer C.
That's the end of this tutorial. Thank you for watching!